Question

a particle is moving in a circular path of radius r. the magnitude of displacement of the particle after covering 1/4 of the circle is​

Answer 1

Answer:

$\sqrt{r^{2} +r^{2} }$

$\sqrt{2r^{2} }$

$r\sqrt{2}$

Step-by-step explanation:

First we take a circle. Then we make it half (semi-circle or 1/2 of the original circle). Now the shortest distance is it's diameter AB because it takes us form point A to point B in the shortest path. Now we divide the semi-circle into half make a quarter circle which is 1/4th of the original circle. So now the shortest distance possible will be AB which joins the two ends of the quarter circle. Now with a bit of manipulation we can see that the line AB is a part of a right angled triangle. So now we can use the formula to calculate the hypotenuse of a right angled triangle which is $A^{2}$ = $B^{2}$ + C. Thus we get the above answer.